显示 XML 源内容

Question

我正在尝试从xml提要中获取内容并将它们显示在列表中。我下面的测试只是为了获得空缺job_title。

$feed = file_get_contents('https://www.jobs.nhs.uk/search_xml?client_id=120650');
$xml = simplexml_load_string($feed);
$items = $xml->nhs_search->vacancy_details;
foreach($items as $item) {
$job_title = $item->job_title;
echo $job_title;
}

下面是 xml 源的片段

<nhs_search>
<vacancy_details>
<id>915854585</id>
<job_title>Band 5 Speech and Language Therapist</job_title>
</vacancy_details>
</nhs_search>

没有显示任何内容，也没有错误。

Answer 1

它工作正常：

$feed = file_get_contents('https://www.jobs.nhs.uk/search_xml?client_id=120650');
$xml = simplexml_load_string($feed);
$items = $xml->vacancy_details;
foreach ($items as $item) {
$id = $item->id;
$job_title = $item->job_title;
echo $id; echo '<br />';
echo $job_title;
}

将$items = $xml->nhs_search->vacancy_details;更改为$items = $xml->vacancy_details;